60：Binary Tree Inorder Traversal

题目：Given a binary tree, return the inorder traversal of its nodes’ values.
For example: Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

有三种解法，分别是递归，用栈迭代以及 Morris 前序遍历

解题代码一：

// 递归版本，时间复杂度 O(n)class Solution {public:        vector<int> inorderTraversal(TreeNode* root) {                vector<int> result;                inorderTraversal(root, result);                return result;        }private:        void inorderTraversal(TreeNode *root, vector<int>& result) {                if (root != nullptr) {                        inorderTraversal(root -> left);                        result.push_back(root -> val);                        inorderTraversal(root -> right);                }        }};

解题代码二：

// 迭代版本，使用栈// 时间复杂度 O(n)，空间复杂度 O(n)class Solution {public:        vector<int> inorderTraversal(TreeNode* root) {                vector<int> result;                stack<const TreeNode*> s;                const TreeNode* cur = root;                while (!s.empty() || cur != nullptr) {                        if (cur != nullptr) {                                s.push(cur);                                cur = cur -> left;                        }                        else {                                cur = s.top();                                s.pop();                                result.push_back(cur -> val);                                cur = cur -> right;                        }                }                return result;        }};

解题代码三：

// morris 遍历// 时间复杂度 O(n)，空间复杂度 O(1)class Solution {public:        vector<int> inorderTraversal(TreeNode* root) {                vector<int> result;                TreeNoede* cur = root;                while (cur != nullptr) {                        if (cur -> left == nullptr) {                                result.push_back(cur -> val);                                cur = cur -> right;                        }                        else {                                TreeNode* node = cur -> left;                                while (node -> right != nullptr && node -> right != cur)                                        node = node -> right;                                if (node -> right == nullptr) {                                        node -> right = cur;                                        cur = cur -> left;                                }                                else {                                        result.push_back(cur -> val);                                        node -> right = nullptr;                                        cur = cur -> right;                                }                        }                }                return result;        }};